Question

how many grams of oxygen are needed to completely burn 35.6 grams of C3H8?

Answer 1

The molar mass of C₃H₈ is
M(C₃H₈) = 3*12g/mol + 8*1g/mol = 36 + 8 = 44 g/mol Theamount of C₃H₈ is
n(C₃H₈) = m(C₃H₈) / M(C₃H₈) = 35.6 g / 44 g/mol = 0.81mol The reaction of burning of C₃H₈ is
C₃H₈ + 5O₂ = 3CO₂ + 4H₂O
According to reaction the amount of oxygen is
n(O₂) = 5*n(C₃H₈) = 5* 0.81 = 4.05 mol
The mass of oxygen is
m(O₂) = n(O₂) * M(O₂) = 4.05 mol * 32 g/mol = 129.6 g